What is the integral of 1/(9x^2+18x+10) ?

The integral of 1/(9x^2+18x+10) is 1/3 arctan(3(x+1))+C

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integral of 1/(9x^2+18x+10)

\int \frac{1}{9 x ^{2} + 18 x + 10} d x

\frac{1}{3} arctan (3 (x + 1)) + C

Solution steps

\int \frac{1}{9 x ^{2} + 18 x + 10} d x

Complete the square

9 x^{2} + 18 x + 10 : 9 (x + 1)^{2} + 1

= \int \frac{1}{9 ( x + 1 ) ^{2} + 1} d x

Apply u-substitution

: \int \frac{1}{9 u ^{2} + 1} d u

= \int \frac{1}{9 u ^{2} + 1} d u

Apply Integral Substitution

= \int \frac{1}{3 ( v ^{2} + 1 )} d v

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= \frac{1}{3} \cdot \int \frac{1}{v ^{2} + 1} d v

Use the common integral:

\int \frac{1}{v ^{2} + 1} d v = arctan (v)

= \frac{1}{3} arctan (v)

Substitute back

= \frac{1}{3} arctan (3 (x + 1))

Add a constant to the solution

= \frac{1}{3} arctan (3 (x + 1)) + C

f (x) = \frac{x + 1}{x - 1}

f (x) = (\frac{3 x - 1}{2 x - 3})

\int_{0}^{1 - ∣ x - 1 ∣} 1 d y

\frac{\partial}{\partial x} (- cos^{2} (x))

What is the integral of 1/(9x^2+18x+10) ?
The integral of 1/(9x^2+18x+10) is 1/3 arctan(3(x+1))+C